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f-R-R-x-i-j-x-R-i-N-j-0-1-f-x-f-i-1-i-1-j-1-x-1-j-0-x-lt-1-x-x-lt-0-f-9-5-




Question Number 1057 by 123456 last updated on 25/May/15
f:R_+ →R  x=i+j  x∈R_+   i∈N  j∈[0,1)  f(x)= { ((f(i−1)+(i+1)(j+1)),(x≥1)),(j,(0≤x<1)),(x,(x<0)) :}  f(9.5)=?
$${f}:\mathbb{R}_{+} \rightarrow\mathbb{R} \\ $$$${x}={i}+{j} \\ $$$${x}\in\mathbb{R}_{+} \\ $$$${i}\in\mathbb{N} \\ $$$${j}\in\left[\mathrm{0},\mathrm{1}\right) \\ $$$${f}\left({x}\right)=\begin{cases}{{f}\left({i}−\mathrm{1}\right)+\left({i}+\mathrm{1}\right)\left({j}+\mathrm{1}\right)}&{{x}\geqslant\mathrm{1}}\\{{j}}&{\mathrm{0}\leqslant{x}<\mathrm{1}}\\{{x}}&{{x}<\mathrm{0}}\end{cases} \\ $$$${f}\left(\mathrm{9}.\mathrm{5}\right)=? \\ $$
Answered by prakash jain last updated on 26/May/15
f(9,5)=f(8)+10∙6  =f(7)+9+10.6  =f(6)+8+9+10.6  =f(5)+7+8+9+10∙6  =f(4)+6+7+8+9+10∙6  =1+2+3+...+9+10+10∙5  =105
$${f}\left(\mathrm{9},\mathrm{5}\right)={f}\left(\mathrm{8}\right)+\mathrm{10}\centerdot\mathrm{6} \\ $$$$={f}\left(\mathrm{7}\right)+\mathrm{9}+\mathrm{10}.\mathrm{6} \\ $$$$={f}\left(\mathrm{6}\right)+\mathrm{8}+\mathrm{9}+\mathrm{10}.\mathrm{6} \\ $$$$={f}\left(\mathrm{5}\right)+\mathrm{7}+\mathrm{8}+\mathrm{9}+\mathrm{10}\centerdot\mathrm{6} \\ $$$$={f}\left(\mathrm{4}\right)+\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}+\mathrm{10}\centerdot\mathrm{6} \\ $$$$=\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{9}+\mathrm{10}+\mathrm{10}\centerdot\mathrm{5} \\ $$$$=\mathrm{105} \\ $$

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